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week8/03-03-2022/03-03-2022.tex

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+%        File: 20-01-2022.tex
+%     Created: 12:28:35 Thu, 20 Jan 2022 EST
+% Last Change: 12:28:35 Thu, 20 Jan 2022 EST
+%
+\documentclass[letterpaper]{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{cancel}
+\usepackage{amssymb}
+\usepackage{listings}
+\usepackage[shortlabels]{enumitem}
+\usepackage{soul}
+%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{mathtools}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{02/17/2022}
+\title{%
+        Induction and Recursion\\
+            \large MATH--190--01 : Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+
+We've seen that $\sum\limits^{100}_{i=0}k=\frac{100*101}{2}=5050$
+Now we're going to find out \emph{why}.
+To do this, we'll use induction.
+This is a demonstration of how induction generally works.
+All of these steps need to explicitly be shown.
+
+First, we need to prove that our predicate is true for one case.
+In this case, we need need to show that $\sum\limits^{1}_{k=1}k=1=\frac{1(1+1)}{2}=1$
+
+Next, we must show that for every $n\geq a$, if $P(n)$ is true, $P(n+1)$ is also true.
+To do this, suppose that the aforementioned case is in fact true. Then show that
+$P(n+1)$ is true. In our case, we assume that $\sum\limits^{n}_{k=1}k=\frac{n(n+1)}{2}$
+is in fact true. Now, we need to prove that it works when $n=n+1$.
+So, we can say the following:
+\begin{equation}
+    \begin{split}
+        \sum^{n+1}_{k=1} k=\sum^{n}_{k=1}k + (n+1)\\
+        =\frac{n(n+1)}{2} +(n+1)\\
+        =\frac{n(n+1)}{2}+\frac{2(n+1)}{2}\\
+        =\frac{n^2+n+2n+2}{2}\\
+        =\frac{n^2+3n+2}{n}\\
+        =\frac{(n+1)(n+2)}{2}
+    \end{split}
+\end{equation}
+
+So, lets prove this:
+\begin{equation}
+    \sum^{n}_{k=0}r^k=\frac{r^{n+1}-1}{r-1}
+\end{equation}
+What must be proved in the base case?
+In this case, we need to prove that it works for $n=0$.
+This is done like so:
+\begin{equation}
+    \sum^{0}_{k=0}r^k=r^0=\frac{r^{0+1}-1}{r-1}=\frac{r-1}{r-1}=1
+\end{equation}
+Next, we state our inductive hypothesis, that is:
+\begin{equation}
+    \sum^{n}_{k=0}r^k=\frac{r^{n+1}-1}{r-1}
+\end{equation}
+Finally, we solve for $n=n+1$:
+\begin{equation}
+    \begin{split}
+        \sum^{n+1}_{k=0}r^k=\sum^{n}_{k=0}r^k+r^{n+1}\\
+        =\frac{r^{n+1}-1}{r-1}+r^{n+1}\\
+        =\frac{r^{n+1}-1}{r-1}+\frac{(r-1)r^{n+1}}{r-1}\\
+        =\frac{\cancel{r^{n+1}}-1+r^{n+2}-\cancel{r^{n+1}}}{r-1}\\
+        =\frac{r^{n+2}-1}{r-1}
+    \end{split}
+\end{equation}
+
+\end{document}